What is the remainder when (100!)100 is divided by 23?

Concept: If an integer N is exactly divisible by a number d, then N leaves remainder 0 when divided by d. Because 0 raised to any positive power is still 0, for any positive integer k, Nk also leaves remainder 0 when divided by d — a number that is a multiple of d stays a multiple of d after being raised to any power.

1. 100! is the product of every integer from 1 to 100: 100! = 1 × 2 × 3 × ... × 22 × 23 × 24 × ... × 100.

2. Since 23 is one of the integers in that product (23 ≤ 100), 23 is a factor of 100!, so 100! is exactly divisible by 23 — that is, 100! leaves remainder 0 when divided by 23.

3. Applying the concept above with N = 100! and d = 23: since 100! leaves remainder 0 when divided by 23, (100!)100 (100! raised to the 100th power) also leaves remainder 0 when divided by 23.

Cross-check: 100! is actually divisible by 23 several times over — it also contains 46 = 2 × 23, 69 = 3 × 23, and 92 = 4 × 23 as factors — so 100! carries a very high power of 23 within it. Raising such a number to the 100th power only increases how many times 23 divides it, confirming the remainder must be 0.